Problem: A circle with radius $3$ has a sector with a $72^\circ$ central angle. What is the area of the sector? ${9\pi}$ $\color{#9D38BD}{72^\circ}$ ${\dfrac{9}{5}\pi}$ ${3}$
Solution: First, calculate the area of the whole circle. Then the area of the sector is some fraction of the whole circle's area. $A_c = \pi r^2$ $A_c = \pi (3)^2$ $A_c = 9\pi$ The ratio between the sector's central angle $\theta$ and $360^\circ$ is equal to the ratio between the sector's area, $A_s$ , and the whole circle's area, $A_c$ $\dfrac{\theta}{360^\circ} = \dfrac{A_s}{A_c}$ $\dfrac{72^\circ}{360^\circ} = \dfrac{A_s}{9\pi}$ $\dfrac{1}{5} = \dfrac{A_s}{9\pi}$ $\dfrac{1}{5} \times 9\pi = A_s$ $\dfrac{9}{5}\pi = A_s$